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3 years ago

What's the geometric mean of 2√3 and 3√3

Mathematics

2 answers:

umka2103 [35]3 years ago

5 0

Answer:

3√2

Step-by-step explanation:

Geometric mean of 2√3 and 3√3 is:

√(2√3 * 3√3) =
√(6*3) =
√(2*9) =
3√2

Dovator [93]3 years ago

3 0

Answer:

4.2

Step-by-step explanation:

Hope this helps

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A machine in the student lounge dispenses coffee. The average cup of coffee is supposed to contain 7.0 ounces. A random sample o

Elenna [48]

Answer:

$t=\frac{7.4-7}{\frac{0.7}{\sqrt{7}}}=1.512$

The degrees of freedom are

$df=n-1=7-1=6$

And the p value for this case is:

$p_v =2*P(t_{(6)}>1.512)=0.181$

The p value is a higher value and using a significance levels of 5% or 10% we see that the p value is higher than the significance level and then we FAIL to reject the null hypothesis and we don't have enough evidence to conclude that the true mean is different from 7 ounces.

Step-by-step explanation:

Information provided

$\bar X=7.4$ represent the sample mean

$s=0.7$ represent the sample standard deviation

$n=7$ sample size

$\mu_o 7$ represent the value to verify

t would represent the statistic

$p_v$ represent the p value

System of hypothesis

We want to verify if the average amount of coffee per cup is different from 7 ounces, the system of hypothesis would be:

Null hypothesis: $\mu = 7$

Alternative hypothesis: $\mu \neq 7$

The statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Replacing the info given we got:

$t=\frac{7.4-7}{\frac{0.7}{\sqrt{7}}}=1.512$

The degrees of freedom are

$df=n-1=7-1=6$

And the p value for this case is:

$p_v =2*P(t_{(6)}>1.512)=0.181$

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A pair of jeans cost $40. They are on sale 20% off. How much will you save?

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$8.00 is how much you will save

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The surface, or rim, of a canyon is at an altitude of 60 m. On a hike down into the canyon, a party of hikers stops for a rest 1

andreyandreev [35.5K]

Start at 60
hike down 137 below so new is 60-137=-77m

then go down again 57
-77-57=-134

new altitude is -134m

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4 years ago

Tyler’s brother earns $12 per hour. The store offers him a raise—a 5% increase per hour. After the raise, how much will Tyler’s

Mumz [18]

Answer:

$12.60 per hour

Step-by-step explanation:

5% of 12 is 0.60

12 + 0.60 = 12.60

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3 years ago

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Let C(x) be the statement "x has a cat," let D(x) be the statement "x has a dog," and let F(x) be the statement "x has a ferret.

jek_recluse [69]

Answer:

$\mathbf{a)} \left( \exists x \in X\right) \; C(x) \; \wedge \; D(x) \; \wedge \; F(x)\\\mathbf{b)} \left( \forall x \in X\right) \; C(x) \; \vee \; D(x) \; \vee \; F(x)\\\mathbf{c)} \left( \exists x \in X\right) \; C(x) \; \wedge \; F(x) \; \wedge \left(\neg \; D(x) \right)\\\mathbf{d)} \left( \forall x \in X\right) \; \neg C(x) \; \vee \; \neg D(x) \; \vee \; \neg F(x)\\\mathbf{e)} \left((\exists x\in X)C(x) \right) \wedge \left((\exists x\in X) D(x) \right) \wedge \left((\exists x\in X) F(x) \right)$

Step-by-step explanation:

Let $X$ be a set of all students in your class. The set $X$ is the domain. Denote

$C(x) - ' \text{$x $ has a cat}'\\D(x) - ' \text{$x$ has a dog}'\\F(x) - ' \text{$x$ has a ferret}'$

$\mathbf{a)}$

Consider the statement 'A student in your class has a cat, a dog, and a ferret'. This means that $\exists x \in X$ so that all three statements C(x), D(x) and F(x) are true. We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as follows

$\left( \exists x \in X\right) \; C(x) \; \wedge \; D(x) \; \wedge \; F(x)$

$\mathbf{b)}$

Consider the statement 'All students in your class have a cat, a dog, or a ferret.' This means that $\forall x \in X$ at least one of the statements C(x), D(x) and F(x) is true. We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as follows

$\left( \forall x \in X\right) \; C(x) \; \vee \; D(x) \; \vee F(x)$

$\mathbf{c)}$

Consider the statement 'Some student in your class has a cat and a ferret, but not a dog.' This means that $\exists x \in X$ so that the statements C(x), F(x) are true and the negation of the statement D(x) . We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as follows

$\left( \exists x \in X\right) \; C(x) \; \wedge \; F(x) \; \wedge \left(\neg \; D(x) \right)$

$\mathbf{d)}$

Consider the statement 'No student in your class has a cat, a dog, and a ferret..' This means that $\forall x \in X$ none of the statements C(x), D(x) and F(x) are true. We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as a negation of the statement in the part a), as follows

$\neg \left( \left( \exists x \in X\right) \; C(x) \; \wedge \; D(x) \; \wedge \; F(x)\right) \iff \left( \forall x \in X\right) \; \neg C(x) \; \vee \; \neg D(x) \; \vee \; \neg F(x)$

$\mathbf{e)}$

Consider the statement ' For each of the three animals, cats, dogs, and ferrets, there is a student in your class who has this animal as a pet.'

This means that for each of the statements C, F and D there is an element from the domain $X$ so that each statement holds true.

We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as follows

$\left((\exists x\in X)C(x) \right) \wedge \left((\exists x\in X) D(x) \right) \wedge \left((\exists x\in X) F(x) \right)$

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4 years ago